Question: Find $\lim_{h\to 0}\dfrac{3(2+h)^4-3(2)^4}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $48$ (Choice B) B $72$ (Choice C) C $96$ (Choice D) D The limit doesn't exist
Answer: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $3(2+h)^4-3(2)^4$, we can tell that the function is $f(x)=3x^4$ and the $x$ -value is $2$. In other words, the limit expression is equal to $f'(2)$ for $f(x)=3x^4$. Let's find $f'(x)$ : $f'(x)=3\cdot4x^3=12x^3$ Now let's evaluate $f'(2)$ : $f'(2)=12(2)^3=96$ In conclusion, $\lim_{h\to 0}\dfrac{3(2+h)^4-3(2)^4}{h}=96$.